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The Grothendieck Group construction seems very similar to the Int construction over traced monoidal categories (pdf).

[It may help to either rewrite the group construction multiplicatively, or to use additive notation for the monoidal structure to make this more obvious]. Roughly speaking the trace is what gives the needed invariance with respect to a common factor on both sides, which is the same thing expressed through quotienting via an equivalence relation in the group construction.

So I must be missing some fundamental difference between these constructions (or I have not read the right paper which already makes this observation). Which is it?

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    $\begingroup$ One difference, the G. group is a group and the Int construction is a category. $\endgroup$
    – Fernando Muro
    Commented Aug 31, 2012 at 21:34
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    $\begingroup$ Aside from Fernando's remark, I'm not sure why you think you're particularly missing anything, because the constructions are indeed similar in spirit. If you think of the category of finite sets and bijections (which forms a traced monoidal category by taking disjoint sum as the monoidal product) as a kind of categorification of he integers, the $Int$ construction gives the symmetric monoidal category of 1-cobordisms, which can be thought of as a kind of categorification of the integers. With a little thought, it might be possible to make the connection tighter. $\endgroup$
    – Todd Trimble
    Commented Aug 31, 2012 at 22:11
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    $\begingroup$ @Todd: 1-cobordism? Are these just plain cobordisms, as per (say) the Wikipedia page? Is this written up anywhere (i.e. that the category of 1-cobordisms can be thought of as a kind of categorification of the integers)? $\endgroup$
    – Jacques Carette
    Commented Aug 31, 2012 at 22:28
  • $\begingroup$ By that, I meant the category whose objects are oriented 0-manifolds and whose morphisms are (oriented diffeo classes of) oriented 1-manifolds with boundary. It's been known a long time that the category 1-Cob is the free rigid symmetric monoidal category on one generator; in the present context, we get it by applying $Int$ to the free symmetric monoidal category on one generator, which is equivalent to the symmetric monoidal groupoid of finite sets and bijections. See Baez-Dolan, HDA 0 (arxiv.org/abs/q-alg/9503002), around page 7. The term "categorification" is being applied [cont.] $\endgroup$
    – Todd Trimble
    Commented Aug 31, 2012 at 23:00
  • $\begingroup$ [cont.] slightly loosely; here the idea is to get the integers as additive group by localizing the symmetric monoidal category 1-Cob at the unit $1 \to x^\ast \otimes x$ and counit $x \otimes x^\ast \to 1$ morphisms (localization being taken in the sense of a symmetric monoidal functor). I'd like to think about this in the context of K-theory a little before I attempt an actual answer (and without guaranteeing that I will answer!). :-) $\endgroup$
    – Todd Trimble
    Commented Aug 31, 2012 at 23:05

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I believe that the Grothendieck group of a commutative monoid can be viewed as a special case of the Int construction. Namely, you can think of a commutative monoid $M$ as a discrete symmetric monoidal category (i.e., a symmetric monoidal category with no non-identity morphisms). This category admits a unique trace, and taking $\operatorname{Int} M$ just gives the Grothendieck group of $M$, thought of as a discrete symmetric monoidal category.

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    $\begingroup$ Well, that's certainly true... $\endgroup$
    – Todd Trimble
    Commented Sep 1, 2012 at 3:09

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